Theorem sb4bor 1764

Description: Simplified definition of substitution when variables are distinct, expressed via disjunction. (Contributed by Jim Kingdon, 18-Mar-2018.)

Assertion

Ref	Expression
sb4bor	|- ( A. x x = y \/ A. x ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) )

Proof of Theorem sb4bor

Step	Hyp	Ref	Expression
1		sb4or 1762	. 2 |- ( A. x x = y \/ A. x ( [ y / x ] ph -> A. x ( x = y -> ph ) ) )
2		sb2 1698	. . . . 5 |- ( A. x ( x = y -> ph ) -> [ y / x ] ph )
3		df-bi 116	. . . . . 6 |- ( ( ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) -> ( ( [ y / x ] ph -> A. x ( x = y -> ph ) ) /\ ( A. x ( x = y -> ph ) -> [ y / x ] ph ) ) ) /\ ( ( ( [ y / x ] ph -> A. x ( x = y -> ph ) ) /\ ( A. x ( x = y -> ph ) -> [ y / x ] ph ) ) -> ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) ) )
4	3	simpri 112	. . . . 5 |- ( ( ( [ y / x ] ph -> A. x ( x = y -> ph ) ) /\ ( A. x ( x = y -> ph ) -> [ y / x ] ph ) ) -> ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) )
5	2, 4	mpan2 417	. . . 4 |- ( ( [ y / x ] ph -> A. x ( x = y -> ph ) ) -> ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) )
6	5	alimi 1390	. . 3 |- ( A. x ( [ y / x ] ph -> A. x ( x = y -> ph ) ) -> A. x ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) )
7	6	orim2i 714	. 2 |- ( ( A. x x = y \/ A. x ( [ y / x ] ph -> A. x ( x = y -> ph ) ) ) -> ( A. x x = y \/ A. x ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) ) )
8	1, 7	ax-mp 7	1 |- ( A. x x = y \/ A. x ( [ y / x ] ph <-> A. x ( x = y -> ph ) ) )

Syntax hints:   -> wi 4   /\ wa 103   <-> wb 104   \/ wo 665   A.wal 1288   [wsb 1693

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 666  ax-5 1382  ax-7 1383  ax-gen 1384  ax-ie1 1428  ax-ie2 1429  ax-8 1441  ax-10 1442  ax-11 1443  ax-i12 1444  ax-4 1446  ax-17 1465  ax-i9 1469  ax-ial 1473  ax-i5r 1474

This theorem depends on definitions:  df-bi 116  df-nf 1396  df-sb 1694

This theorem is referenced by: (None)